Question

Evaluate
$3\frac{\sin {72}^o}{\cos {18}^o}-\frac{\sec {32}^o}{cosec{58}^o}$

Answer 1

Given $3\frac{\sin {72}^{\circ }}{\cos {18}^{\circ }}-\frac{\sec {32}^{\circ }}{cosec{58}^{\circ }}$

$\sin {72}^{\circ }=\sin ({90}^{\circ }-{18}^{\circ })$

$=\cos \left({18}^{\circ }\right)$

$\sec {32}^{\circ }=\frac{1}{\cos {32}^{\circ }}$

$cosec{58}^{\circ }=\frac{1}{\sin {58}^{\circ }}$

Then

$\Rightarrow 3.\frac{\cos {18}^{\circ }}{\cos {18}^{\circ }}-\frac{1/\cos {32}^{\circ }}{1/\sin {58}^{\circ }}$

$\Rightarrow 3.\frac{\cos {18}^{\circ }}{\cos {18}^{\circ }}-\frac{\sin {58}^{\circ }}{\cos {32}^{\circ }}$

$\Rightarrow \sin {58}^{\circ }=\sin ({90}^{\circ }-{32}^{\circ })$

$=\cos \left({32}^{\circ }\right)$

$\Rightarrow 3.\frac{\cos {18}^{\circ }}{\cos {18}^{\circ }}-\frac{\cos {32}^{\circ }}{\cos {32}^{\circ }}$

$=3-1$

$=2$